2.3 Volumes of Revolution: Cylindrical Shells

Big idea

Slicing a solid of revolution perpendicular to its axis produces disks and washers, and that works whenever you can write the boundary curves as functions of the variable running along the axis. Sometimes you cannot, or you can only at the cost of solving an equation you would rather not solve, or of splitting the region into pieces because the boundary switches curves halfway up. This section gives the other slicing direction.

Cut the plane region into thin strips running parallel to the axis of rotation instead of perpendicular to it, and spin one strip. A thin rectangle at distance $x$ from a vertical axis, with height $h$ and width $dx$, sweeps out a cylindrical tube of radius $x$, height $h$ and wall thickness $dx$. Slit that tube down one side and unroll it: it flattens into a slab of length $2\pi x$, width $h$ and thickness $dx$, so its volume is about $2\pi x h\, dx$. Add the shells and let them thin out, and you get $2\pi \int x h(x)\, dx$.

So this is the same three-step move as every other section in this chapter: slice, size one slice, integrate. What differs is the direction of the cut. Shells and washers always agree when both apply; they organize the same volume differently, and one of the two setups is usually much shorter to write down.

The rule of thumb worth learning: shells win when you are rotating about a line parallel to the strips’ long direction. Rotate about a vertical line and vertical strips sweep out shells, so you integrate in $x$ and never have to solve the curve for $x$ as a function of $y$. Rotate about a horizontal line and horizontal strips do the same job, with the variables exchanged. Whenever inverting the boundary curve looks unpleasant, or the region has a single formula in one variable and two in the other, reach for shells.

Decoder

Rotating about a vertical line, a thin vertical strip of the region sweeps out a cylindrical shell whose volume is approximately the circumference times the height times the thickness, and the total volume is the integral of $2\pi$ times radius times height.

The unrolling argument is what makes the approximation legitimate. A shell of inner radius $x$, outer radius $x + \Delta x$ and height $h$ has exact volume $\pi h \big[(x + \Delta x)^2 - x^2\big] = 2\pi x h \,\Delta x + \pi h (\Delta x)^2$. The leftover term carries $(\Delta x)^2$, so it vanishes faster than the main term and contributes nothing to the limit.

Two quantities must be identified separately in every problem, and confusing them is the standard error. The radius is the distance from the axis to the strip, which equals $x$ only when the axis is the $y$-axis. The height is the extent of the region along the strip, which is the top curve minus the bottom curve, not simply $f(x)$ unless the region rests on $y = 0$.

One more distinction: in the washer method you integrate along the axis of rotation, in the shell method perpendicular to it. So the limits in a shell problem are the smallest and largest radii, not the range of heights.

Definitions and results

The shell element. A cylindrical shell of radius $r$, height $h$ and thickness $dr$ has volume approximately $2\pi r h\, dr$: circumference times height times thickness.

Shells about the $y$-axis. If the region under $y = f(x)$ and above $y = 0$ for $0 \le a \le x \le b$ is rotated about the $y$-axis,

$$ V = 2\pi \int_a^b x\, f(x)\, dx $$

Shells between two curves. If the region is bounded above by $f$ and below by $g$ on $[a, b]$ with $a \ge 0$, the strip height is the gap:

$$ V = 2\pi \int_a^b x \big[ f(x) - g(x) \big] \, dx $$

Shells about the $x$-axis. Exchange the roles of the variables. Horizontal strips at height $y$, with horizontal extent $p(y) - q(y)$, give

$$ V = 2\pi \int_c^d y \big[ p(y) - q(y) \big] \, dy $$

Rotating about another vertical line. The radius is the distance from the axis to the strip. For the line $x = k$ with the region lying to the right of it, the radius is $x - k$; for a region lying to the left, it is $k - x$. In both cases the radius is $|x - k|$ and it must be nonnegative across the whole interval.

Shells versus washers. About a vertical line, shells integrate in $x$ and washers in $y$; about a horizontal line, the reverse. Choose the one whose boundary curves are already solved for the needed variable and that avoids splitting the region.

When shells are clearly better. If the region’s upper boundary is a curve like $y = 2x - x^2$ that requires the quadratic formula to invert, or a curve like $y = x e^{x}$ that cannot be inverted in elementary terms, rotating about a vertical line should be done with shells.

Both methods agree. Shells and washers compute the same number whenever both apply. Running the second method is a genuine check on the first, and worth doing while you are learning the setup.

Worked examples

A basic shell, checked with washers

Rotate the region bounded by $y = x^2$, $y = 0$ and $x = 2$ about the $y$-axis.

Vertical strips run parallel to the axis of rotation. A strip at position $x$ has radius $x$ and height $x^2$, and $x$ ranges from $0$ to $2$:

$$ V = 2\pi \int_0^2 x \cdot x^2 \, dx = 2\pi \left[ \frac{x^4}{4} \right]_0^2 = 2\pi \cdot 4 = 8\pi $$

Check with washers. Slicing perpendicular to the $y$-axis, a horizontal slice at height $y$ runs from the curve $x = \sqrt{y}$ out to the line $x = 2$, and $y$ runs from $0$ to $4$:

$$ V = \pi \int_0^4 \big[ 2^2 - (\sqrt{y})^2 \big] dy = \pi \int_0^4 (4 - y)\, dy = \pi \left[ 4y - \frac{y^2}{2} \right]_0^4 = \pi(16 - 8) = 8\pi $$

The two agree, and notice how much more setup the washer version needed: solving for $x$, finding the height range, and seeing that the outer radius is constant.

A region whose boundary resists inverting

Rotate the region between $y = 2x - x^2$ and the horizontal axis about the $y$-axis.

The parabola meets the axis where $x(2 - x) = 0$, so $x$ runs from $0$ to $2$. With shells the setup is immediate: radius $x$, height $2x - x^2$.

$$ V = 2\pi \int_0^2 x(2x - x^2)\, dx = 2\pi \int_0^2 (2x^2 - x^3)\, dx = 2\pi \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2 $$

$$ = 2\pi \left( \frac{16}{3} - 4 \right) = 2\pi \cdot \frac{4}{3} = \frac{8\pi}{3} $$

With washers you would first solve $y = 2x - x^2$ for $x$, getting the two branches $x = 1 \pm \sqrt{1 - y}$, then integrate $\pi\big[(1 + \sqrt{1-y})^2 - (1 - \sqrt{1-y})^2\big]$ from $0$ to $1$. That is the same answer through more algebra, and it is the situation the shell method exists for.

Rotating about a vertical line off the axis

Rotate the region under $y = x^2$ from $x = 0$ to $x = 1$ about the line $x = 2$.

The region lies to the left of the axis of rotation, so the radius of the strip at position $x$ is $2 - x$, which is positive for every $x$ in $[0, 1]$. The height is still $x^2$.

$$ V = 2\pi \int_0^1 (2 - x)\, x^2 \, dx = 2\pi \int_0^1 (2x^2 - x^3)\, dx = 2\pi \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^1 $$

$$ = 2\pi \left( \frac{2}{3} - \frac{1}{4} \right) = 2\pi \cdot \frac{5}{12} = \frac{5\pi}{6} $$

A quick plausibility check: the region has area $\int_0^1 x^2 dx = \tfrac{1}{3}$ and its typical point sits roughly $1.5$ units from the axis, so the swept volume should be near $2\pi(1.5)\left(\tfrac{1}{3}\right) = \pi$. The exact answer $\tfrac{5\pi}{6}$ is close to that estimate, as it should be.

Shells about a horizontal axis

Rotate the region bounded by $x = y^2$, $x = 0$ and $y = 1$ about the $x$-axis.

Here the strips must run horizontally to be parallel to the axis. A strip at height $y$ has radius $y$ and horizontal extent from $x = 0$ to $x = y^2$, so its length is $y^2$, and $y$ runs from $0$ to $1$:

$$ V = 2\pi \int_0^1 y \cdot y^2 \, dy = 2\pi \left[ \frac{y^4}{4} \right]_0^1 = \frac{\pi}{2} $$

Check with washers, slicing vertically. At position $x$ in $[0, 1]$, the region runs from $y = \sqrt{x}$ up to $y = 1$, so the outer radius is $1$ and the inner radius is $\sqrt{x}$:

$$ V = \pi \int_0^1 \big[ 1 - x \big] dx = \pi \left[ x - \frac{x^2}{2} \right]_0^1 = \frac{\pi}{2} $$

The methods agree again, which is the reassurance you want while the setup still feels unfamiliar.

Practice

These problems all rotate a plane region about a line and ask for the volume by shells. For each one, name the axis, decide which direction the strips must run to be parallel to it, then write the radius as a distance and the height as a difference of curves before integrating.

Quiz

Five items on setting up shell integrals, choosing shells over washers, and rotating about lines other than the coordinate axes.